Engineering Mechanics.pdf

CHAPTER 2
ENGINEERING MECHANICS
GATE Previous Year Solved Paper For Mechanical Engineering
Published by: NODIA and COMPANY ISBN: 9788192276250
Visit us at: www.nodia.co.in
YEAR 2012 TWO MARKS
Common Data For Q.
• 1 and 2
Two steel truss members, and
AC BC , each having cross sectional area of
100 mm2
, are subjected to a horizontal force F as shown in figure. All the
joints are hinged.
MCQ 2.1 If 1 kN
F = , the magnitude of the vertical reaction force developed at the
point B in kN is
(A) 0.63 (B) 0.32
(C) 1.26 (D) 1.46
MCQ 2.2 The maximum force F is kN that can be applied at C such that the axial
stress in any of the truss members DOES NOT exceed 100 MPa is
(A) 8.17 (B) 11.15
(C) 14.14 (D) 22.30
YEAR 2011 ONE MARK
MCQ 2.3 The coefficient of restitution of a perfectly plastic impact is
(A) 0 (B) 1
(C) 2 (D) 3
MCQ 2.4 A stone with mass of 0.1 kg is catapulted as shown in the figure. The total
force Fx (in N) exerted by the rubber band as a function of distance x

ENGINEERING MECHANICS CHAP 2
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(in m) is given by F x
300
x
2
= . If the stone is displaced by 0.1 m from the
un-stretched position ( )
x 0
= of the rubber band, the energy stored in the
rubber band is
(A) 0.01 J (B) 0.1 J
(C) 1 J (D) 10 J
YEAR 2011 TWO MARKS
MCQ 2.5 A 1 kg block is resting on a surface with coefficient of friction 0.1
μ = . A
force of 0.8 N is applied to the block as shown in the figure. The friction
force is
(A) 0 (B) 0.8 N
(C) 0.98 N (D) 1.2 N
YEAR 2009 ONE MARK
MCQ 2.6 A block weighing 981 N is resting on a horizontal surface. The coefficient of
friction between the block and the horizontal surface is 0.2
μ = . A vertical
cable attached to the block provides partial support as shown. A man can
pull horizontally with a force of 100 N. What will be the tension, T (in N)
in the cable if the man is just able to move the block to the right ?
(A) 176.2 (B) 196.0

CHAP 2 ENGINEERING MECHANICS PAGE 73
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(C) 481.0 (D) 981.0
YEAR 2009 TWO MARKS
MCQ 2.7 A uniform rigid rod of mass M and length L is hinged at one end as shown
in the adjacent figure. A force P is applied at a distance of /
L
2 3 from the
hinge so that the rod swings to the right. The reaction at the hinge is
(A) P
− (B) 0
(C) /
P 3 (D) /
P
2 3
YEAR 2008 ONE MARK
MCQ 2.8 A straight rod length ( )
L t , hinged at one end freely extensible at the other
end, rotates through an angle ( )
t
θ about the hinge. At time t, ( ) 1
L t =
m, ( ) 1
L t =
o m/s, ( )
t
4
θ π
= rad and ( ) 1
t
θ =
o rad/s. The magnitude of the
velocity at the other end of the rod is
(A) 1 m/s (B) 2 m/s
(C) 3 m/s (D) 2 m/s
YEAR 2008 TWO MARKS
MCQ 2.9 A circular disk of radius R rolls without slipping at a velocity V . The
magnitude of the velocity at point P (see figure) is

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(A) V
3 (B) /2
V
3
(C) /2
V (D) 2 /
V 3
MCQ 2.10 Consider a truss PQR loaded at P with a force F as shown in the figure -
The tension in the member QR is
(A) 0.5 F (B) 0.63 F
(C) 0.73 F (D) 0.87 F
YEAR 2007 ONE MARK
MCQ 2.11 During inelastic collision of two particles, which one of the following is
conserved ?
(A) Total linear momentum only
(B) Total kinetic energy only
(C) Both linear momentum and kinetic energy
(D) Neither linear momentum nor kinetic energy
YEAR 2007 TWO MARKS
MCQ 2.12 A block of mass M is released from point P on a rough inclined plane with
inclination angle θ, shown in the figure below. The co-efficient of friction is
μ. If tan
<
μ θ, then the time taken by the block to reach another point Q
on the inclined plane, where PQ s
= , is
(A)
( )
cos tan
g
s
2
θ θ μ
−
(B)
( )
cos tan
g
s
2
θ θ μ
+

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(C)
( )
sin tan
g
s
2
θ θ μ
−
(D)
( )
sin tan
g
s
2
θ θ μ
+
YEAR 2006 TWO MARKS
MCQ 2.13 If a system is in equilibrium and the position of the system depends upon
many independent variables, the principles of virtual work states that the
partial derivatives of its total potential energy with respect to each of the
independent variable must be
(A) 1.0
− (B) 0
(C) 1.0 (D) 3
MCQ 2.14 If point A is in equilibrium under the action of the applied forces, the values
of tensions TAB and TAC are respectively
(A) 520 N and 300 N (B) 300 N and 520 N
(C) 450 N and 150 N (D) 150 N and 450 N
YEAR 2005 ONE MARK
MCQ 2.15 The time variation of the position of a particle in rectilinear motion is given
by 2 2
x t t t
3 2
= + + . If v is the velocity and a is the acceleration of the
particle in consistent units, the motion started with
(A) 0, 0
v a
= = (B) 0, 2
v a
= =
(C) 2, 0
v a
= = (D) 2, 2
v a
= =
MCQ 2.16 A simple pendulum of length of 5 m, with a bob of mass 1 kg, is in simple
harmonic motion. As it passes through its mean position, the bob has a
speed of 5 m/s. The net force on the bob at the mean position is
(A) zero (B) 2.5 N
(C) 5 N (D) 25 N
YEAR 2005 TWO MARKS
MCQ 2.17 Two books of mass 1 kg each are kept on a table, one over the other. The

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coefficient of friction on every pair of contacting surfaces is 0.3. The lower
book is pulled with a horizontal force F . The minimum value of F for which
slip occurs between the two books is
(A) zero (B) 1.06 N
(C) 5.74 N (D) 8.83 N
MCQ 2.18 A shell is fired from a cannon. At the instant the shell is just about to leave
the barrel, its velocity relative to the barrel is 3 m/s, while the barrel is
swinging upwards with a constant angular velocity of 2 rad/s. The magnitude
of the absolute velocity of the shell is
(A) 3 m/s (B) 4 m/s
(C) 5 m/s (D) 7 m/s
MCQ 2.19 An elevator (lift) consists of the elevator cage and a counter weight, of mass
m each. The cage and the counterweight are connected by chain that passes
over a pulley. The pulley is coupled to a motor. It is desired that the elevator
should have a maximum stopping time of t seconds from a peak speed v. If
the inertias of the pulley and the chain are neglected, the minimum power
that the motor must have is
(A) mV
2
1 2
(B)
t
mV
2
2
(C)
t
mV2
(D)
t
mV
2 2
MCQ 2.20 A 1 kg mass of clay, moving with a velocity of 10 m/s, strikes a stationary
wheel and sticks to it. The solid wheel has a mass of 20 kg and a radius of

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1 m. Assuming that the wheel is set into pure rolling motion, the angular
velocity of the wheel immediately after the impact is approximately
(A) zero (B)
3
1 rad/s
(C)
3
10 rad/s (D)
3
10 rad/s
YEAR 2004 ONE MARK
MCQ 2.21 The figure shows a pin-jointed plane truss loaded at the point M by hanging
a mass of 100 kg. The member LN of the truss is subjected to a load of
(A) 0 Newton (B) 490 Newtons in compression
(C) 981 Newtons in compression (D) 981 Newtons in tension
YEAR 2004 TWO MARKS
MCQ 2.22 An ejector mechanism consists of a helical compression spring having a
spring constant of 981 10
k N/m
3
#
= . It is pre-compressed by 100 mm from
its free state. If it is used to eject a mass of 100 kg held on it, the mass will
move up through a distance of

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(A) 100 mm (B) 500 mm
(C) 581 mm (D) 1000 mm
MCQ 2.23 A rigid body shown in the figure (a) has a mass of 10 kg. It rotates with
a uniform angular velocity ‘ω’. A balancing mass of 20 kg is attached as
shown in figure (b). The percentage increase in mass moment of inertia as a
result of this addition is
(A) 25% (B) 50%
(C) 100% (D) 200%
MCQ 2.24 The figure shows a pair of pin-jointed gripper-tongs holding an object
weighting 2000 N. The coefficient of friction (μ) at the gripping surface
is 0.1 XX is the line of action of the input force and YY is the line of
application of gripping force. If the pin-joint is assumed to be frictionless,
the magnitude of force F required to hold the weight is
(A) 1000 N (B) 2000 N
(C) 2500 N (D) 5000 N

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YEAR 2003 ONE MARK
MCQ 2.25 A truss consists of horizontal members (AC,CD, DB and EF) and vertical
members (CE and DF) having length l each. The members AE, DE and BF
are inclined at 45c to the horizontal. For the uniformly distributed load “p”
per unit length on the member EF of the truss shown in figure given below,
the force in the member CD is
(A)
pl
2
(B) pl
(C) 0 (D)
pl
3
2
MCQ 2.26 A bullet of mass “m” travels at a very high velocity v (as shown in the
figure) and gets embedded inside the block of mass “M ” initially at rest on
a rough horizontal floor. The block with the bullet is seen to move a distance
“s” along the floor. Assuming μ to be the coefficient of kinetic friction
between the block and the floor and “g” the acceleration due to gravity
what is the velocity v of the bullet ?
(A)
m
M m gs
2μ
+ (B)
m
M m gs
2μ
−
(C)
( )
m
M m
gs
2
μ
μ
+
(D)
m
M gs
2μ
YEAR 2003 TWO MARKS
Common Data For Q.
• Data for Q. 27 & 28 are given below. Solve
the problems and choose correct answers.
A reel of mass “m” and radius of gyration “k” is rolling down smoothly from
rest with one end of the thread wound on it held in the ceiling as depicated
in the figure. Consider the thickness of thread and its mass negligible in
comparison with the radius “r ” of the hub and the reel mass “m”. Symbol
“g” represents the acceleration due to gravity.

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MCQ 2.27 The linear acceleration of the reel is
(A)
( )
r k
gr
2 2
2
+
(B)
( )
r k
gk
2 2
2
+
(C)
( )
r k
grk
2 2
+
(D)
( )
r k
mgr
2 2
2
+
MCQ 2.28 The tension in the thread is
(A)
( )
r k
mgr
2 2
2
+
(B)
( )
r k
mgrk
2 2
+
(C)
( )
r k
mgk
2 2
2
+
(D)
( )
r k
mg
2 2
+
YEAR 2001 ONE MARK
MCQ 2.29 A particle P is projected from the earth surface at latitude 45c with escape
velocity 11.19 /
km s
v = . The velocity direction makes an angle α with the
local vertical. The particle will escape the earth’s gravitational field
(A) only when 0
α = (B) only when 45c
α =
(C) only when 90c
α = (D) irrespective of the value of α
MCQ 2.30 The area moment of inertia of a square of size 1 unit about its diagonal is
(A)
3
1 (B)
4
1
(C)
12
1 (D)
6
1

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YEAR 2001 TWO MARKS
MCQ 2.31 For the loading on truss shown in the figure, the force in member CD is
(A) zero (B) 1 kN
(C) kN
2 (D)
2
kN
1
MCQ 2.32 Bodies 1 and 2 shown in the figure have equal mass m. All surfaces are
smooth. The value of force P required to prevent sliding of body 2 on body
1 is
(A) 2 mg
P = (B) mg
P 2
=
(C) 2 mg
P 2
= (D) P mg
=
MCQ 2.33 Mass M slides in a frictionless slot in the horizontal direction and the bob
of mass m is hinged to mass M at C , through a rigid massless rod. This
system is released from rest with 30c
θ = . At the instant when 0c
θ = , the
velocities of m and M can be determined using the fact that, for the system
(i.e., m and M together)

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(A) the linear momentum in x and y directions are conserved but the energy
is not conserved.
(B) the linear momentum in x and y directions are conserved and the energy
is also conserved.
(C) the linear momentum in x direction is conserved and the energy is also
conserved.
(D) the linear momentum in y direction is conserved and the energy is also
conserved.
*********

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SOLUTION
SOL 2.1 Option (A) is correct.
From above figure. Three forces are acting on a common point. Hence by
Lami’s Theorem.
( )
sin
F
105c sin sin
T T
120 135
2 1
c c
= =
&
sin
T
135
1
c sin sin
F
105 105
1
c c
= =
T1 0.7320 kN
=
Hence vertical reaction at B,
RNT1
cos
T 30
1 c
= . cos
0 73205 30
# c
= 0.634 kN
=
SOL 2.2 Option (B) is correct.
From Previous question
sin
F
105c sin
T
120
2
c
=
T2 .
sin
sin F F
135
120 0 8965
#
c
= =
and T1 ( . )F
0 73205
=
T2 T
> 1
σ 100 MPa
= (given)
As we know F A1
σ #
=
& Fmax A
max 1
σ #
=
T2 100 100
#
=
. F
0 8965 100 100
#
=
F
.
11154.5 N
0 8965
100 100
#
= = 11.15 kN
=

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From the Newton’s Law of collision of Elastic bodies.
Velocity of separation e #
= Velocity of approach
( )
V V
2 1
− ( )
e U U
1 2
= −
Where e is a constant of proportionality & it is called the coefficient of
restitution and its value lies between 0 to 1. The coefficient of restitution
of a perfectly plastic impact is zero, because all the K.E. will be absorbed
during perfectly plastic impact.
Given : Fx x
300 2
= , Position of x is, x 0
= to .
x 0 1
=
The energy stored in the rubber band is equal to work done by the stone.
Hence dE F dx
x
=
Integrating both the sides & put the value of F & limits
dE
E
0
# x dx
300
.
2
0
0 1
= #
E x
300
3
.
3
0
0 1
= : D 300
( . )
0.1 Joule
3
0 1 3
= =
; E
Given : 1 kg
m = , μ .
0 1
= ; From FBD : RN mg
=
Now static friction force,
fS R mg
N
μ μ
= = 0.1 1 9.8 0.98 N
# #
= =
Applied force F 0.8 N
= is less then, the static friction 0.98 N
fS =
F f
< S
So, we can say that the friction developed will equal to the applied force
F 0.8 N
=
SOL 2.6 Option (C) is correct.
Given : 981 N
W = , 0.2
μ =
First of all we have to make a FBD of the block
Here, RN = Normal reaction force

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T =Tension in string
Using the balancing of forces, we have
0
Fx
Σ = : RN
μ 100 N
=
RN
.
500 N
100
0 2
100
μ
= = =
and 0
Fy
Σ = or downward forces = upward forces
W T RN
= + & T W RN
= − 981 500 48 N
1
= − =
When rod swings to the right, linear acceleration a and angular acceleration
α comes in action. Centre of gravity (G) acting at the mid-point of the rod.
Let R be the reaction at the hinge.
Linear acceleration a .
r α
= L
2 # α
=
L
a
2
= ...(i)
and about point G, for rotational motion
G
Μ
/ IG # α
=
R L P L
2 6
+
b b
l l
ML
L
a
12
2
2
= b l From equation (i)
R P
3
+ Ma
3
=
a
M
R
M
P
3
= + ...(ii)

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By equilibrium of forces in normal direction to the rod
Fm
/ 0
= : P R
− Ma
= M
M
R
M
P
3
= +
b l From equation (ii)
P R
− R P
3
= +
& R 0
= So, reaction at the hinge is zero.
SOL 2.8 Option (D) is correct.
Let : Vt =Tangential Velocity
Vr =Relative Velocity
V =Resultant Velocity
Let rod of length ( )
L t increases by an amount ( )
L t
T .
Given ( )
L t 1
= m, ( )
L t
:
=1 m/sec, ( )
t
4
θ π
= rad, ( )
t
θ
:
=1 rad/sec
Time taken by the rod to turn
4
π rad is,
t
( )
( )
tan
velocity
dis ce
t
t
θ
θ
= = :
/
1
4
4
π π
= = sec
So, increase in length of the rod during this time will be
( )
L t
Δ ( )
L t t
#
=
4
1
4
#
π π
= = meter
Rod turn
4
π radian. So, increased length after
4
π sec, (New length)
.
1
4
1 785
π
= + =
a k m
Now, tangential velocity,
Vt .
R ω
= . .
1 785 1 1 785
#
= = m/sec ( )
t
ω θ
= o
Radial velocity, Vr ( )
L t
=
:
=1 m/sec

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Therefore, the resultant velocity will be
VR V V
t r
2 2
= + ( . ) ( ) 2.04 2 /sec
m
1 785 1
2 2
-
= + =
When disc rolling along a straight path, without slipping. The centre of the
wheel O moves with some linear velocity and each particle on the wheel
rotates with some angular velocity.
Thus, the motion of any particular on the periphery of the wheel is a
combination of linear and angular velocity.
Let wheel rotates with angular velocity=ω rad/sec.
So, ω
R
V
= ....(i)
Velocity at point P is, VP PQ
#
ω
= ...(ii)
From triangle OPQ PQ ( ) ( ) 2 ( )
cos
OQ OP OQ OP POQ
2 2
+
# #
+ −
( ) ( ) 2 120
cos
R R RR
2 2
c
= + −
( ) ( ) ( )
R R R
2 2 2
= + + R
3
= ...(iii)
From equation (i), (ii) and (iii)
VP
R
V R
3
#
= V
3
=
The forces which are acting on the truss PQR is shown in figure.
We draw a perpendicular from the point P, that intersects QR at point S .

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Let PS QS a
= =
RQ & RR are the reactions acting at point Q & R respectively.
Now from the triangle PRS
tan30c
SR
PS
= & SR 1.73
tan
PS a a a
30
3
3
1
c
= = = =
Taking the moment about point R,
( . )
R a a
1 73
Q # + .
F a
1 73
#
=
RQ
.
.
.
. .
F F F
a
a
2 73
1 73
2 73
1 73 0 634
= = =
From equilibrium of the forces, we have
R R
R Q
+ F
=
RR 0.634 F
F R F
Q
= − = − 0.366 F
=
To find tension in QR we have to use the method of joint at point Q, and
F 0
y
Σ =
sin
F 45
QP c RQ
=
FQP
0.634 0.8966
F F
2
1
= =
and, 0
Fx
Σ =
cos
F 45
QP c FQR
= & FQR 0.8966 0.634 0.63
F F F
2
1 -
#
= =
In both elastic & in inelastic collision total linear momentum remains
conserved. In the inelastic collision loss in kinetic energy occurs because the
coefficient of restitution is less than one and loss in kinetic energy is given
by the relation,
. .
K E
T
( )
( ) ( )
m m
m m u u e
2
1
1 2
1 2
1 2
2 2
=
+
− −
First of all we resolve all the force which are acting on the block.

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Given : PQ s
= where N = Normal fraction force
μ tan
< θ
Now from Newton’s second law,
F ma
=
sin
mg N
θ μ
− ma
= Acceleration of block
a =
sin cos
mg mg
θ μ θ
− ma
= cos
N mg θ
=
sin cos
g g
θ μ θ
− a
=
a cos
cos
sin
g θ
θ
θ μ
= −
: D ( )
cos tan
g θ θ μ
= − ...(i)
From the Newton;s second law of Motion,
s ut at
2
1 2
= + ( )
cos tan
g t
0
2
1 2
θ θ μ
= + − u 0
=
t
( )
cos tan
g
s
2
θ θ μ
=
−
If a system of forces acting on a body or system of bodies be in equilibrium
and the system has to undergo a small displacement consistent with the
geometrical conditions, then the algebraic sum of the virtual works done by
all the forces of the system is zero and total potential energy with respect to
each of the independent variable must be equal to zero.
First we solve this problem from Lami’s theorem. Here three forces are
given. Now we have to find the angle between these forces
Applying Lami’s theorem, we have
sin
F
90c sin sin
T T
120 150
AB AC
c c
= =
1
600
/ /
T T
3 2 1 2
AB AC
= =
TAB 600 300 520 N
2
3 3 .
#
= =
TAC 300 N
2
600
= =

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Alternative :
Now we using the Resolution of forces.
Resolve the TAB & TAC in x & y direction (horizontal & vertical components)
We use the Resolution of forces in x & y direction
0
Fx
Σ = , cos
T 60
AB c cos
T 30
AC c
=
T
T
AC
AB
2
3
1
2 3
#
= = ...(i)
0
Fy
Σ = , sin sin
T T
60 30
AB AC
c c
+ 600 N
=
T T
2
3
2
1
AB AC
+ N
600
=
T T
3 AB AC
+ 1200 N
= T T
3
AC
AB
= From equation (i)
Now, T T
3
3
AB
AB
+ 1200 N
=
T
4 AB 1200 3
=
TAB 520 N
4
1200 3
= =
and TAC 300 N
T
3 3
520
AB
= = =
Given ; x t t t
2 2
3 2
= + +
We know that,
v
dt
dx
= ( )
dt
d t t t
2 2
3 2
= + + t t
6 2 2
2
= + + ...(i)
We have to find the velocity & acceleration of particle, when motion stared,
So at t 0
= , v 2
=
Again differentiate equation (i) w.r.t. t
a 12 2
dt
dv
dt
d x t
2
2
= = = +
At t 0
= , a 2
=

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We have to make the diagram of simple pendulum
Here, We can see easily from the figure that tension in the string is balanced
by the weight of the bob and net force at the mean position is always zero.
Given : m m
1 2
= 1 kg
= , μ .
0 3
=
The FBD of the system is shown below :
For Book (1) 0
Fy
Σ = RN1 mg
= ...(i)
Then, Friction Force FN1 R mg
N1
μ μ
= =
From FBD of book second,
0
Fx
Σ = , F R R
N N
1 2
μ μ
= +
0
Fy
Σ = , RN2 2 mg
R mg mg mg
N1
= + = + = ...(ii)
For slip occurs between the books when
F R R
N N
1 2
$ μ μ
+ 2 mg
mg
$ μ μ #
+
F (3 )
mg
$ μ . ( . )
0 3 3 1 9 8
# #
$ .
8 82
$
It means the value of F is always greater or equal to the 8.82, for which slip

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occurs between two books.
So, F 8.83 N
=
Given : ω 2 /sec
rad
= , r 2 m
=
Tangential velocity of barrel, Vt rω
= 2 2
#
= 4 /sec
m
=
V V V
i j
r t
= + 3 4
i j
= +
Resultant velocity of shell, V ( ) ( )
3 4
2 2
= + 5 /sec
m
25
= =
Given : Mass of cage & counter weight = m kg each
Peak speed V
=
Initial velocity of both the cage and counter weight.
V1 /sec
m
V
=
Final velocity of both objects
V2 0
=
Initial kinetic Energy, E1 mV mV mV
2
1
2
1
2 2 2
= + =
Final kinetic Energy E2 ( ) ( )
m m
2
1 0
2
1 0 0
2 2
= + =
Now, Power = Rate of change of K.E.
t
E E
1 2
= −
t
mV2
=
Given : m1 1 kg
= , 10 /sec
m
V1 = , m2 20 kg
= , V2 = Velocity after striking
the wheel 1
r = meter
Applying the principal of linear momentum on the system
dt
dP 0
= P
& = constant
Initial Momentum = Final Momentum
m V
1 1
# ( )
m m V
1 2 2
= +
V2
( )
m m
m V
1 20
1 10
21
10
1 2
1 1 #
=
+
=
+
=

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Now after the collision the wheel rolling with angular velocity ω.
So, V2 rω
= & ω 0.476
r
V
21 1
10
2
#
= = =
It is nearly equal to /
1 3.
First of all we consider all the forces, which are acting at point L.
Now sum all the forces which are acting along x direction,
FLK FLM
= Both are acting in opposite direction
Also summation of all the forces, which are acting along y-direction.
FLN 0
= Only one forces acting in y-direction
So the member LN is subjected to zero load.
Given : 981 10
k 3
#
= N/m, x 100
i = mm=0.1 m, m 100
= kg
Let, when mass m 100
= kg is put on the spring then spring compressed by
x mm. From the conservation of energy :
Energy stored in free state = Energy stored after the mass is attach.
( . .)
K E i ( . .) ( . .)
K E P E
f f
= +
kx
2
1
i
2
( 0.1)
kx mg x
2
1 2
= + +
kxi
2
2 ( 0.1)
kx mg x
2
= + +
Substitute the values, we get
( . )
981 10 0 1
3 2
# # (981 10 ) [2 100 9.81 ( 0.1)]
x x
3 2
# # # # #
= + +
10 10
3 2
#
−
( . )
x x
10 2 0 1
3 2
= + +
10 .
x x
1000 2 0 2
2
= + +
.
x x
1000 2 9 8
2
+ − 0
=
Solving above equation, we get
x
( ) ( . )
2 1000
2 2 4 1000 9 8
2
#
! #
=
− − −
2000
2 4 39200
2000
2 198
! !
= − + = −
On taking -ve sign, we get

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x
2000
2 198
10
1
= − − =− , 100 mm
m =−
(-ve sign shows the compression of the spring)
Given : First Mass, m1 10 kg
=
Balancing Mass, m2 kg
20
=
We know the mass moment of inertia, I mk2
=
Where, k = Radius of gyration
Case (I) : When mass of 10 kg is rotates with uniform angular velocity ‘ω’
I1 m k
1 1
2
= 10 (0.2)2
#
= 10 0.04 0.4 kg m2
#
= = 0.2 m
k1 =
Case (II) : When balancing mass of 20 kg is attached then moment of inertia
I2 ( . ) ( . )
10 0 2 20 0 1
2 2
# #
= + . . .
0 4 0 2 0 6
= + = Here 0.2 m
k1 =
and 0.1 m
k2 =
Percent increase in mass moment of inertia,
I 100
.
. . 100
I
I I
0 4
0 6 0 4
1
2 1
# #
= − = − 100 50%
2
1
#
= =
Given : Weight of object W 2000 N
=
Coefficient of Friction μ 0.1
=
First of all we have to make the FBD of the system.
Here, RN = Normal reaction force acting by the pin joint.
F RN
μ
= = Friction force
In equilibrium condition of all the forces which are acting in y direction.
R R
N N
μ μ
+ 2000 N
=
RN
μ 1000 N
=
RN
.
10000 N
0 1
1000
= = .
0 1
μ =
Taking the moment about the pin, we get
10000 150
# F 300
#
=
F 5000 N
=

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Given : AC CD DB EF CE DF l
= = = = = =
At the member EF uniform distributed load is acting, the U.D.L. is given
as “p” per unit length.
So, the total load acting on the element EF of length l
= Lord per unit length # Total length of element
p l pl
#
= =
This force acting at the mid point of EF . From the FBD we get that at A
and B reactions are acting because of the roller supports, in the upward
direction. In equilibrium condition,
Upward force = Downward forces
R R
a b
+ pl
= ...(i)
And take the moment about point A,
pl l l
2
# +
b l ( )
R l l l
b
= + +
pl l
2
3
# R l
3
b #
= & Rb
pl
2
=
Substitute the value of Rb in equation (i), we get
R
pl
2
a + pl
=
Ra pl
pl pl
2 2
= − = R
pl
2
b
= =
At point A we use the principal of resolution of forces in the y-direction,
F 0
y =
/ : sin
F 45
AE c R
pl
2
a
= =
FAE
sin
pl pl pl
2 45
1
2
2
2
c
# #
= = =
And FAC 45
cos
F
pl pl
2 2
1
2
AE c #
= = =
At C , No external force is acting. So,
FAC
pl
F
2 CD
= =

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Given : Mass of bullet m
=
Mass of block M
=
Velocity of bullet v
=
Coefficient of Kinematic friction = μ
Let, Velocity of system (Block + bullet) after striking the bullet u
=
We have to make the FBD of the box after the bullet strikes,
Friction Force (Retardation) Fr
=
Applying principal of conservation of linear momentum,
dt
dP 0
= or P mV
= .
tan
cons t
=
So, mv ( )
M m u
= +
u
M m
mv
=
+
...(i)
And, from the FBD the vertical force (reaction force),
RN ( )
M m g
= +
Fr ( )
R M m g
N
μ μ
= = +
Frictional retardation a
( )
( )
m M
F
M m
M m g
r μ
=
+
− =
+
− +
g
μ
=− ...(ii)
Negative sign show the retardation of the system (acceleration in opposite
direction). From the Newton’s third law of motion,
Vf
2
2
u as
2
= +
Vf = Final velocity of system (block + bullet) = 0
2
u as
2
+ 0
=
u2
as
2
=− ( )
g s
2 # #
μ
=− − gs
2μ
= From equation (ii)
Substitute the value of u from equation (i), we get
M m
mv 2
+
a k gs
2μ
=
( )
M m
m v
2
2 2
+
gs
2μ
=
v2 ( )
m
gs M m
2
2
2
μ
=
+

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v gs
m
M m
2 #
μ
= +
b l m
M m gs
2μ
= +
Given : Mass of real m
=
Radius of gyration k
=
We have to make FBD of the system,
Where, T =Tension in the thread
mg = Weight of the system
Real is rolling down. So Angular acceleration ( )
α comes in the action
From FBD, For vertical translation motion,
mg T
− ma
= ...(i)
and for rotational motion,
MG
Σ IG α
=
T r
# mk
r
a
2
#
= I mk
G
2
= , /
a r
α =
T
r
mk a
2
2
#
= ...(ii)
From equation (i) & (ii) Substitute the value of T in equation (i), we get
mg
r
mk a
2
2
#
− ma
=
mg a
r
mk m
2
2
= +
; E
a
k r
gr
2 2
2
=
+
...(iii)
From previous question, T mg ma
= −
Substitute the value of a from equation (iii), we get
T
( )
mg m
k r
gr
2 2
2
#
= −
+ ( )
( )
k r
mg k r mgr
2 2
2 2 2
=
+
+ −
k r
mgk
2 2
2
=
+

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We know that a particle requires the velocity of 11.2 /
km s for escape it from
the earth’s gravitational field. The angle α does not effect on it.
The BD is the diagonal of the square ABCD and CBD 45c
= .
From the BCE
T sin45c
BC
CE
= & CE 1 45
sin
2
1
c
#
= = unit
where CE is the height of the triangle BCD
T .
Now, the area moment of inertia of a triangle about its base BD is bh
12
3
where b = base of triangle & h = height of triangle
So, the triangle ABD
T are same and required moment of inertia of the
square ABCD about its diagonal is,
I 1 ( ) ( )
BD CE
2
12
3
# # #
=
6
1 2
2
1
12
1
3
# #
= =
c m unit
The reactions at the hinged support will be in only vertical direction as
external loads are vertical.
Now, consider the FBD of entire truss. In equilibrium of forces.
R R
a f
+ 1 1 2 kN
= + = ...(i)
Taking moment about point A, we get
R L
3
f # 1 1 2 3
L L L
# #
= + =
Rf 1 kN
=
From equation (i), Ra 2 1 1 kN
= − =

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First consider the FBD of joint A with the direction of forces assumed in
the figure.
Resolving force vertically, we get
Ra sin
F 45
AB c
=
FAB
sin
kN
45
1 2
c
= = (Compression)
Resolving forces horizontally
FAC cos
F 45
AB c 1 kN
2
2
1
#
= = (Tension)
Consider the FBD of joint B with known value of force FAB in member AB
Resolving forces vertically,
FBC cos
F 45
AB c
= 1 kN
2
2
1
#
= = (Tension)
Resolving forces horizontally,
FBD sin
F 45
AB c
= 1 kN
2
2
1
#
= = (Compression)
Consider the FBD of joint C with known value of force FBC and FAC
Resolving forces vertically,
1 sin
F F 45
BC CD c
= +

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1 sin
F F
1 45 0
CD CD
&
c
= + =
From the FBD of the system.
RN cos
mg 45c
=
All surfaces are smooth, so there is no frictional force at the surfaces.
The downward force sin
mg 45c is balanced by cos
P 45c.
sin
mg 45c cos
P 45c
=
mg
2
1
# P
2
1
#
= & P mg
=
In this case the motion of mass m is only in x -direction. So, the linear
momentum is only in x -direction & it remains conserved.
Also from Energy conservation law the energy remains constant i.e. energy
is also conserved.
**********

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